The polynomial equation of degree 4 having real coefficients with three of its roots as \(2 \pm \sqrt{3}\)…
The polynomial equation of degree 4 having real coefficients with three of its roots as \(2 \pm \sqrt{3}\) and \(1+2 i\), is
\(x^4-6 x^3-14 x^2+22 x+5=0\)
\(x^4-6 x^3-19 x+22 x-5=0\)
\(x^4-6 x^3+19 x-22 x+5=0\)
\(x^4-6 x^3+14 x^2-22 x+5=0\)
Solution
It is given that, the polynomial equation of degree 4 having real coefficients with three of its roots as \(2 \pm \sqrt{3}\) and \(l+2 i\), so the remaining root is \(1-2 i\). Now, the quadratic equation whose roots as \(2 \pm \sqrt{3}\) is
\(x^2-4 x+1=0 \text {, and }\)
the quadratic equation whose roots as \(1 \pm 2 i\), is
\(x^2-2 x+5=0\)
So, the required polynomial equation is
\(\begin{aligned}
\left(x^2-4 x+1\right)\left(x^2-2 x+5\right) & =0 \\
\Rightarrow \quad x^4-6 x^3+14 x^2-22 x+5 & =0
\end{aligned}\)
Hence, option (4) is correct.